a = 27.2/.158 = 172 radians/second^2
18*2pi= 0 + 0t + (1/2)(172)t^2
113 = 86 t^2
t = 1.15 seconds
w = 0 + a t
w = 172*1.15 = 197 radians/second
A constant torque of 27.2 N · m is applied to
a grindstone for which the moment of inertia
is 0.158 kg · m2.
Find the angular speed after the grindstone
has made 18 rev.
Assume the grindstone
starts from rest.
Answer in units of rad/s
1 answer
1 answer